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Let $C$ be a projective curve (not necessarily reduced or irreducible). Let $\mathcal{F}, \mathcal{G}$ be $\mathcal{O}_C$-modules and $\phi:\mathcal{F} \to \mathcal{G}$ be a morphism of $\mathcal{O}_C$-modules. Suppose further that $\mathcal{F}$ is locally free. Is it true that the kernel of $\phi$ is also locally free?

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You ask if every submodule of a locally free module $F$ is again locally free. Of course this fails. First of all, this submodule doesn't have to be quasi-coherent. But even when it is quasi-coherent, and we therefore may pass to affine covers, it fails even for $F=\mathcal{O}$, i.e. not every quasi-coherent ideal is locally free. For this it would have to be zero or invertible.

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If you assume the curve smooth, then yes. This is because for a smooth curve, the local rings $\mathcal O_{X,x}$ are DVRs or fields, e.g. PIDs. In particular, a submodule of a free module over a PID is again free. Therefore, the exact sequence $$ 0 \to \ker \varphi \to \mathcal F \to \mathcal G $$ can be seen at the stalks as the sequence $$ 0 \to \ker \varphi_x \to \mathcal O_{X,x}^{\oplus I} \to \mathcal G_x. $$ where $I$ is some index set. Therefore $\ker \varphi_x$ is free for each $x \in X$, which means $\ker \varphi$ is locally free.

Of course not all curves are smooth, so with non-smooth curves you get counter-examples.

Another situation where the answer is "yes" is when $\varphi : \mathcal F \to \mathcal G$ is surjective and $\mathcal G$ is locally free. This is because looking at the exact sequence at the stalks, since $\mathcal G_x$ is free, it is projective, thus the exact sequence splits ; in particular, $\ker \varphi_x$ is projective. But a projective module over a local ring is free, so that $\ker \varphi_x$ is free for all $x \in X$.

Hope that helps,

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